alleryFor water, K_b = 0.52 K kg mol⁻¹. What is the boiling point of a solution of 18 g glucose (M = 180) in 1 kg water?
A373.20 K
B373.67 K
C374.71 K
D373.15 K
Answer & Solution
Correct answer: A. 373.20 K
We have m = (18/180)/1.
So m = 0.1 mol kg⁻¹.
ΔT_b = 0.52 × 0.1 = **0.052 K**. New b.p.
= 373.15 + 0.052 = **373.20 K**.
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